6.2: Linear Functions on Hyperplanes

It is not always so easy to write a linear operator as a matrix. Generally, this will amount to solving a linear systems problem. Examining a linear function whose domain is a hyperplane is instructive.

Example 63

Let $$V=\left\{ c_{1}\begin{pmatrix}1\\1\\0\end{pmatrix} +c_{2}\begin{pmatrix}0\\1\\1\end{pmatrix} \middle| c_{1},c_{2}\in \Re \right\} $$ and consider \(L:V\to \Re^{3}\) defined by
$$
L\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix},\qquad
L\begin{pmatrix}0\\1\\1\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix}.
$$
By linearity this specifies the action of \(L\) on any vector from \(V\) as
$$
L\left[ c_{1}\begin{pmatrix}1\\1\\0\end{pmatrix} + c_{2} \begin{pmatrix}0\\1\\1\end{pmatrix} \right]= (c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}.
$$
The domain of \(L\) is a plane and its range is the line through the origin in the \(x_{2}\) direction. It is clear how to check that \(L\) is linear.

It is not clear how to formulate \(L\) as a matrix;
since
\begin{eqnarray*}
L\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =
\begin{pmatrix}
0&0&0\\
1&0&1\\
0&0&0
\end{pmatrix}
\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =(c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,
\end{eqnarray*}
or since
\begin{eqnarray*}
L\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =
\begin{pmatrix}
0&0&0\\
0&1&0\\
0&0&0
\end{pmatrix}
\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =(c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}
\end{eqnarray*}
you might suspect that \(L\) is equivalent to one of these \(3\times3\) matrices. It is not. All \(3\times3\) matrices have \(\Re^{3}\) as their domain, and the domain of \(L\) is smaller than that. When we do realize this \(L\) as a matrix it will be as a \(3\times2\) matrix. We can tell because the domain of \(L\) is 2 dimensional and the codomain is \(3\) dimensional